## Maharashtra State Board Class 8 Maths Solutions Chapter 15 Area Practice Set 15.5

Question 1.

Find the areas of given plots. (All measures are in meters.)

Solution:

i. Here, ∆QAP, ∆RCS are right angled triangles and ☐QACR is a trapezium.

In ∆QAP, l(AP) = 30 m, l(QA) = 50 m

A(∆QAP)

= \(\frac { 1 }{ 2 }\) x product of sides forming the right angle

= \(\frac { 1 }{ 2 }\) x l(AP) x l(QA)

= \(\frac { 1 }{ 2 }\) x 30 x 50

= 750 sq. m

In ☐QACR, l(QA) = 50 m, l(RC) = 25 m,

l(AC) = l(AB) + l(BC)

= 30 + 30 = 60 m

A(☐QACR)

= \(\frac { 1 }{ 2 }\) x sum of lengths of parallel sides x height

= \(\frac { 1 }{ 2 }\) x [l(QA) + l(RC)] x l(AC)

= \(\frac { 1 }{ 2 }\) x (50 + 25) x 60

= \(\frac { 1 }{ 2 }\) x 75 x 60

= 2250 sq.m

In ∆RCS, l(CS) = 60 m, l(RC) = 25 m A(∆RCS)

= \(\frac { 1 }{ 2 }\) x product of sides forming the right angle

= \(\frac { 1 }{ 2 }\) x l(CS) x l(RC)

= \(\frac { 1 }{ 2 }\) x 60 x 25

= 750 sq. m

In ∆PTS, l(TB) = 30 m,

l(PS) = l(PA) + l(AB) + l(BC) + l(CS)

= 30 + 30 + 30 + 60

= 150m

A(∆PTS) = \(\frac { 1 }{ 2 }\) x base x height

= \(\frac { 1 }{ 2 }\) x l(PS) x l(TB)

= \(\frac { 1 }{ 2 }\) x 150 x 30

= 2250 sq. m

∴ Area of plot QPTSR = A(∆QAP) + A(☐QACR) + A(∆RCS) + A(∆PTS)

= 750 + 2250 + 750 + 2250

= 6000 sq. m

∴ The area of the given plot is 6000 sq.m.

ii. In ∆ABE, m∠BAE = 90°, l(AB) = 24 m, l(BE) = 30 m

∴ [l(BE)]² = [l(AB)]² + [l(AE)]²

…[Pythagoras theorem]

∴ (30)² = (24)² + [l(AE)]²

∴ 900 = 576 + [l(AE)]²

∴ [l(AE)]² = 900 – 576

∴ [l(AE)]² = 324

∴ l(AE) = √324 = 18 m

…[Taking square root of both sides]

A(∆ABE)

= \(\frac { 1 }{ 2 }\) x product of sides forming the right angle

= \(\frac { 1 }{ 2 }\) x l(AE) x l(AB)

= \(\frac { 1 }{ 2 }\) x 18 x 24

= 216 sq. m

In ∆BCE, a = 30m, b = 28m, c = 26m

∴ Area of plot ABCDE

= A(∆ABE) + A(∆BCE) + A(∆EDC)

= 216 + 336 + 224

= 776 sq. m

∴ The area of the given plot is 776 sq.m.

[Note: In the given figure, we have taken l(DF) = 16 m]